# Application of Lagrange rule.how is functioning Lagrange rule if the function is f(x)=(x+1)^-1*(x+2)^-1 and the interval is [1;2]?

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### 1 Answer

We'll re-write the function:

f(x) = 1/(x+1)(x+2)

The function is continuous over the interval [1;2] (the function is discontinuously for x = -1 and x = -2, that are the roots of the denominator). Also, the function, being continuous, it could be differentiated over the range [1; 2].

We'll apply Lagrange's theorem over the range [1; 2]:

f(2) - f(1) = f'(c)(2 - 1), if c is in the range (1 ; 2).

f'(c) = [f(2) - f(1)]/(2-1)

f(2) = 1/(2+1)(2+2)

f(2) = 1/12

f(1) = 1/(1+1)(1+2)

f(1) = 1/6

f(2) - f(1) = 1/12 - 1/6

f(2) - f(1) = (1-2)/12

f(2) - f(1) = -1/12

Now, we'll determine f'(x):

f'(x) = [1/(x+1)(x+2)]'

f'(x) = [1/(x^2 + 3x + 2)]'

f'(x) = (-2x - 3)/[(x+1)(x+2)]^2

f'(c) = (-2c - 3)/[(c+1)(c+2)]^2

But, according to Lagrange's theorem, we'll get:

f'(c) = -1/312

f'(c) = (-2c - 3)/[(c+1)(c+2)]^2

(-2c - 3)/[(c+1)(c+2)]^2 = -1/12

12(2c+3) = [(c+1)(c+2)]^2

24c + 36 = [(c+1)(c+2)]^2

**The number c, that is in the opened interval (1 ; 2), has to verify the identity **

**24c + 36 = [(c+1)(c+2)]^2,**

**for Lagrange's theorem to be valid.**